Appendix BMock Test 1 Section A: Single Correct Answer Type 1. An elevator is accelerating upwards with an acceleration of 6 m/s2. Inside it, a person of mass 50 kg is standing on a weighing machine which is kept on an inclined plane having the angle of inclination 60°. The reading of the weighing machine is (c) Point 9 on the string has the greatest downward acceleration (d) Point 2 on the string has a downward velocity and upward acceleration. 9 1 2 Pv t a = 6 m/s Le a di Fig. B.2 ge The center of mass moves towards left. The center of mass moves towards right. The center of mass remains stationary. The net torque about the center of mass of the spool is zero. 3. The rate of dissipation of heat by a black body at a temperature T is Q. The rate of dissipation of heat by another body at temperature 2T and emissivity 0.25 is (a) 16Q (b) 4Q (c) 8Q (d) 8.5Q 4. Figure B.3 shows a sinusoidal wave of period T travelling to the right along a string at time t = 0. Which of the following statements is incorrect? (a) Point 3 on the string is moving upward with maximum speed. (b) Point 5 on the string has the greatest upward acceleration. ga In Rough I. + II. + + + III. + + + P � P � P P � � IV. + The correct order of choices in a decreasing order of magnitude of force on P is (a) II > I > III > IV (b) I > II > III > IV (c) II > I > IV > III (d) III > IV > I > II 6. Find the equivalent resistance across AB. A C en 5. Given are four arrangements of three fixed electric charges. In each arrangement, a point labeled P is also identified—test charge, +q, is placed at point P. All of the charges are of the same magnitude Q, but they can be either positive or negative as indicated. The charges and point P all lie on a straight line. The distances between adjacent items, either between two charges or between a charge and point P, are all the same. ar ni F Fig. B.3 ng (a) 40 kg (b) 160 kg (c) 80 kg (d) 50 kg 2. A spool is pulled horizontally by two equal and opposite forces as shown in Fig. B.2. Which of the following statements is correct? (a) (b) (c) (d) 6 5 Fig. B.1 7 4 Weighing machine F 8 3 2 60° .L td . Paper 1 2W 2W 2W 2W B Fig. B.4 2W B.2 Physics loop shown in the figure? The integration in done in the sense shown. The loop has N turns and a part of helical loop on which arrows are drawn is outside the plane of paper. ga ge Le ar ni ng (a) –m0(NI) (b) m0(I) (c) m0(NI) (d) Zero 8. A flag is mounted on a car moving due north with the velocity of 20 km/h. Strong winds are blowing due east with the velocity of 20 km/h. The flag will point in direction (a) East (b) North-East (c) South-East (d) South-West 9. Water coming out of the mouth of a tap of area of cross-section 2.5 cm2 is falling vertically in a streamline flow with the speed 3 m/s. The area of cross-section of the water column 80 cm below the tap is In Fig. B.5 di a I 11. For a particle moving in straight line with increasing speed, the appropriate sign of acceleration a and velocity v can be (a) a > 0 and v > 0 (b) a < 0 and v < 0 (c) a > 0 and v < 0 (d) a < 0 and v > 0 12. A number of forces of different magnitudes and directions which are variable in nature are used to move a particle along a smooth curved horizontal path. (a) The work done on the particle by the resultant force equals the change in the kinetic energy of particle. (b) No work done is possible on a particle in circular motion. (c) Work done is possible on a particle in circular motion but it will not be equal to change in KE. (d) If the speed of particle changes during circular motion, then some work is being done on the particle. 13. A particle of mass m = 4 kg moving at 6i m/s col- .L td . Ú Section B: Multiple Correct Answers Type Pv t (a) 1 W (b) 2 W (c) 3 W (d) 4 W 7. In Fig. B.5, a wire carries current I. What is the value B · dl (as in Ampere’s law) on the helical of the Fig. B.6 1f / cm V0 (a) 0.5 cm (b) 1 cm2 2 (c) 1.5 cm (d) 2 cm2 10. A particle is executing S.H.M. A and B are two extreme positions in which its velocity is zero. It passes through a certain point P at intervals of 0.5 s and 1.5 s with a speed of 3 m/s. What is the maximum speed of the particle? en (b) 2 2 m/s (c) 4 2 m/s (d) 3 2 m/s 2f / cm Volts Metal 1 C (a) 6 2 m/s (a) Velocity of the center of mass (CM) is 5i m/s. (b) The velocities of the particles relative to the center of mass have same magnitude. (c) The speed of individual particle before and after collision remains same. (d) The velocity of particles relative to CM after collision are v = - i m/s, v = 2i m/s 14. The graph between the stopping potential (V0) and wave number (1/l) is as shown in Fig. B.7. If f is the work function, then 2 1 lides perfectly elastically with a particle of mass m2 = 2 kg moving at 3i m/s. Metal 2 f1 f2 f3 q q q 0.001 0.002 0.004 Metal 3 (1/nm) 1/l Fig. B.7 (a) f1 : f2 : f3 = 1 : 2 : 4 (b) f1 : f2 : f3 = 4 : 2 : 1 (c) tan q µ hc / e, where q is the slope. Appendix B: Mock Tests B.3 .L td . an input power of 16 W produced a rise of 10 K in the liquid. When the power was doubled, the same temperature rise was achieved by making the rate of flow of liquid three times faster. Find the power lost (in W) to the surrounding in each case. 18. A block of mass m = 1 kg is projected on a smooth horizontal floor with a speed v0 = 3 m/s towards a fixed light spring of stiffness k = 16 N/m. The time of contact of the smooth block with the spring is p sec. Find the value of ‘*’. * Pv t (d) Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only. 15. A wave is transmitted from medium 1 to medium 2 and the respective velocities in the two media are v1 and v2, respectively, (a) The phase of the transmitted wave does not change if v2 > v1. (b) The amplitude of transmitted wave is always less than that of the incident wave. (c) The frequency of the transmitted wave is always equal to that of the incident wave. (d) The phase of the transmitted wave does not change if v1 < v2. v0 k m a Fig. B.9 In 19. A string 120 cm in length sustains a transverse standing wave. The points of the string where the displacement amplitude is each 3.5 mm are separated by the distance of 15 cm. Find the overtone in which the string vibrates. 20. In the circuit shown in Fig. B.10, each battery is 5 V and has an internal resistance of 0.2 W. The reading of the ideal voltmeter is V. Find V in volt. ar ni ng 16. A cubical vessel with a liquid of density r = 103 kg/m3 is kept at rest on an inclined plane of angle of inclination q = 37°. If b = 1 m, then find the pressure difference between A and B. di Section C: Integer Answer Type A B b v Le q Fig. B.8 ge 17. In two experiments with a continuous flow calorimeter to determine the specific heat capacity of a liquid, ga Paper 2 Fig. B.10 Q P V Section A: One or More Correct Answers Type C en 1. A metallic square loop PQRS is moving in its own plane with velocity v in a uniform magnetic field perpendicular to its plane as shown in Fig. B.11. If VP, VQ, VR and VS are the potentials of points P, Q, R and S then which of the following is an incorrect statement? (a) VP = VQ (b) VP > VS (c) VP > VR (d) VS > VR S R Fig. B.11 2. A bomb of mass 3m is kept inside a closed box of mass 3m and length 4L at its center. It explodes in two parts of mass m and 2m. The two parts move in opposite directions and stick to the opposite sides of the walls of box. The box is kept on a smooth horizontal surface. then the induced charge on its outer surface is uniformly distributed. There are two coils A and B as shown in Fig.13 (a) When switch S is closed. . Which of the followings is/are correct? ng 3. (d) When switch S is opened. B. What maximum deviation a ray of same wavelength can undergo at the interface of two media when entering from medium (2)? (a) 90° (b) 45° (c) 0° (d) 60° 5. The deviation for a ray at the interface of two media from denser (1) to rarer (2) with the angle of incidence 30° is 15°. the direction of the momentary induced current in coil B will be in clockwise direction.B. the direction of the momentary current induced in coil B will be in clockwise direction.v22 = È 2h ˘ (c) v1 + v2 = Í (n1 . 8. Pick the correct statements: (a) If a point charge is placed off-center inside an electrically neutral spherical metal shell. If the velocities of the photoelectrons (of mass m) coming out are v1 and v2. B. (d) v12 + v22 = C en w m C Fig. ar ni 1/ 2 È 2h ˘ (a) v1 . (b) If a point charge is placed off-center inside an electrically neutral.12 What is the distance moved by the box during this time interval? A L (d) 3 + � S a Fig. the direction of the momentary current induced in coil B will be in anticlockwise direction.v2 = Í (n1 .n 2 )˙ Îm ˚ Le O 1/ 2 l ga ge 2h (n1 .n 2 ) m 6. An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. A small ball of mass m suspended from the ceiling at a point O by a thread of length moves along a horizontal circle with a constant angular velocity w. (b) When switch S is closed. there is no electrostatic force on the particle due to the shell. (d) Magnitude of angular momentum about O is constant.13. the direction of the momentary induced current in coil B will be in anticlockwise direction. 4. respectively.n 2 ) m 2h (n1 . isolated spherical metal shell. The average power delivered by the source (a) must be 1000 W (b) may be greater than 1000 W (c) may be less than 1000 W (d) All of the above three are possible. B. B. then the induced charge on its inner surface is uniformly distributed. . 7.14 (a) Angular momentum is constant about O.4 Physics Fig. (d) If a charged particle is located inside a nonmetal shell of uniform charge. then B di (a) 0 L (c) 12 In L (b) 6 Pv t 4L (c) A non-metal shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charges were concentrated at the center of the shell. Two identical photocathodes receive the light of frequencies v1 and v2.L td . (c) Vertical component of angular momentum about O is constant. (c) When switch S is opened.n 2 )˙ m Î ˚ (b) v12 . (b) Angular momentum is constant about C. current in the inductor is zero. 16. The magnitude of maximum magnetic field is Pv t Section B: Comprehension Type r R Fig. For Problems 9–11 As shown in Fig. B y C R O • Fig.Appendix B: Mock Tests B. pulleys are light and smooth. Friction is only between block C and floor. B. Area under V–t graph represents (a) total charge flown (b) change in current (c) product of inductance and change in current (d) product of inductance and total charge flown Maximum value of current in the inductor will take place at time t equal to (a) 4 s (b) 2 s (c) 0 (d) 1 s 17. J = J0[1 – (r/R)]. (a) 5 Mg sin q/8 (b) Mg sin q/4 (c) Mg sin q/6 (d) Mg sin q/8 For Problems 12–14 An infinite cylindrical wire of radius R and having current density varying with its radius r as. Graph between the magnetic field and radius is B A 2 M/ (a) B (b) a q . Find the tension in the string. B. 10. (a) > g sin q (b) < g sin q (c) g sin q (d) Any of the above can be possible Regarding the accelerations of A and C. we can say that: (a) accelerations of both will be same (b) acceleration of A will be greater than that of C (c) acceleration of C will be greater than that of A (d) any of the above is possible. Then answer the following questions. At t = 0. 11.15 di (c) R B R r R r (d) r For Problems 15–19 An inductor of inductance 3 H is given across which a potential difference varying with time is shown in Fig.17. System is released from rest. In 9.15. B. B.L td . The position where magnetic field strength is maximum is (a) R (b) 3R/4 (c) R/2 (d) R/4 13. Match the following columns I and II. 12. blocks of masses M/2. (a) 1 A (b) 3 A (c) 5 A (d) 7 A Section C: Matching Column Type 18. Find the current in the circuit at t = 3 s. B.17 15.16 x V (volt) 12 0 2 4 t (s) Fig. Heat given to process is positive.5 C 2 M/ B tan q m= 2 M (a) J0R m0 6 (b) 3 J 0 Rm0 16 (c) 5 J 0 Rm0 16 (d) 5 J 0 Rm0 6 14. M and M/2 are connected through a light string as shown. . • en ga ge Le ar ni ng Find the acceleration of block B. If R is the radius of the earth. The ball collides elastically with the blocks. where ve is the escape velocity and k < 1. A small ball is projected horizontally between two large blocks. B. A battery of emf E is connected across a conductor as shown in Fig. If the velocity of the blocks do not change due to the collision. The square can rotate about point . the object distance u = 40 ± 0.6 Physics r1 E 30 20 r2 H E G F Fig. As one observes from A to B. then find out the velocity of the ball after the second collision.75 (c) 0. The maximum % error in the measurement of f is (a) 1.k2 R en ga ge 1. Assume friction to be absent Fig.19 Fig. Entries of column I are to be matched with entries of column II. In the measurement of the focal length f of a concave mirror.3 (d) 2. B.18 ar ni ng Column I Column II (p) EF (a) DW < 0 (q) FG (b) DW > 0 (r) GH (c) DQ < 0 (s) HE (d) DQ > 0 19.k2 (d) R 1+ k2 2. B. Column II Column I (p) Torque (a) Planck’s constant (q) Angular Momentum (b) Work (r) Latent heart (c) Strain (s) Angle (d) Gravitatinal Potential Pv t 3 V (m ) a 20 di 10 In 10 B A .75 (b) 0. Match quantities in column I and with those quantities in column II having same dimensions.21.20 (a) 5V (b) 7V (c) 9V (d) None of these 4.L td . Four forces of the same magnitude act on a square as shown in Fig. B. Column I Column II (p) Current (a) increases (q) Drift velocity (b) decreases of electron (r) Electric field (c) remains same (s) Potential drop (d) cannot be deter per unit length mine 20. Column I and column II contain four entries each. B. P (atm) Le Paper 1 Section A: Single Correct Answer Type (a) 1.1 cm. The ball is given a velocity V m/s and each of the large blocks move uniformly with a velocity of 2V m/s.B. match the following. the maximum height to which it will rise measured from the center of earth will be (neglect air resistance)? C (c) R (1–k)2 (b) R 1.19.2 cm.25 Mock Test 2 3. A projectile is fired vertically upwards from the surface of the earth with a velocity Kve. the image distance v = 20 ± 0. During the fall. A uniform magnetic field B = 3i + 4 j + k exists in region of space. the work done by one mole of gas during heating from T0 to hT0 through the given process will be: F2 O 2 (i + j + k ) di F3 (a) In q en y Fig. B.i + j + k ) (a) Ê h . A container filled with mercury and having a wooden block floating in it is allowed to fall freely under gravity. The total heat generated in the circuit after the closing of switch is C (d) 2 (.k ) ar ni R E (c) ng -q0 Pv t F1 Fig. 0) is placed in x-y plane as shown in Fig. B. The force on semicircular wire will be (a) (1/2)CE (c) (1/4)CE2 R/2 ga ge . B. 1 ln h a Le 2 2 (i + j . A capacitor having initial charge q0 = CE/2 is connected to a cell of emf E as shown in Fig. B. A semicircular wire of radius 1 m carrying current 1 A having its center at (2.1˜¯ (c) a ln h – (g – 1) RT0 (d) None of these 9.O. The molar heat capacity C for an ideal gas going through a given process is given by C = a/T.25.22 (b) (1/8)CE2 (d) None of these 7.Á RT0 Ë g . B. If g = CP /CV.24 1m C (2. It slides a distance d along the track in time t to . The voltage of an AC source varies with time according to the relation: E = 120 sin 100 pt cos 100 pt V.22. A box of mass m is released from rest at position 1 on the frictionless curved track shown in Fig.j + k ) a F4 (b) 8. 2) (a) q/e0 (c) q/4e0 (b) q/2e0 (d) zero Section B: Multiple Correct Answer Type 45° x Fig. B.21 +q0 2 (i .23 11.23. What is the peak voltage of the source? (a) 60 V (b) 120 V (c) 30 V (d) V 6. where a is a constant. B.L td . Flux passing through the shaded surface of sphere when a point charge q is placed at the center is (radius of the sphere is R) Fig. 2.1ˆ (b) a ln h . the upthrust on the wooden block will be (a) zero (b) equal to the weight of mercury displaced by the immersed portion of the block (c) equal to the weight of the block in air (d) equal to the loss of weight of the block in water 10. The force which can produce greatest torque is (a) F1 (b) F2 (c) F3 (d) F4 5. mid point of one of the edges. The Sun’s mass is about 3. then a d 16. What is her y-coordinate (in m) if the same wave front reaches her at the same instant as it does the first two listeners? 20. Then: (a) if the compression is isothermal. (c) if the compression is adiabatic. then their final volumes will be same.2 ¥ 105 times the Earth’s mass. A third listener is positioned along the positive y-axis. An elastic metal rod will change its length when it (a) falls vertically under its weight (b) is pulled along its length by a force acting at one end (c) rotates about an axis at one end (d) slides on a rough surface .B. A point source of sound is located somewhere along the x-axis. 14. Find the ratio of the magnitudes of the gravitational pull of the Sun on the Moon (Fms) and of the Earth on the Moon (Fme). then their final volumes will be different. Paper 2 .8 Physics 1 Section C: Integer Answer Type Fig. then the monoatomic gas will have maximum final volume. (b) if the compression is adiabatic. B. 17. respectively. Pv t 2 di h In m 15. 18. Experiments show that the same wave front simultaneously reaches listeners at x = –8 m and x = +2.25 ge Le ar ni ng (a) h = vt (b) h = (1/2)gt2 2 (c) d = (1/2)at (d) mgh = (1/2)mv2 12. B. reach position 2. The distance of instantaneous axis of rotation from the lowest plate is equal to R /*. The sun is about 400 times as far from the Earth as the Earth is from the Moon. Assume that the Sun-Moon distance is constant and equal to SunEarth distance. The particle takes t1 seconds to go from A to M and t2 seconds to go from M to B. Let v and a be the instantaneous speed and instantaneous acceleration. respectively. Each is compressed till their pressure becomes twice the initial pressure. Find the value of capacitor in mF in L–C circuit.26 19. respectively. A body rolls on two horizontal plates 1 and 2 moving with velocities 3v and –v. (d) Maximum speed of resultant SHM will be more than double of the initial SHMs. 13.L td . In Young’s double slit experiment. The inductor in an L–C oscillation has a maximum potential difference of 10 V and maximum energy of 100 mJ. Its velocities at A and B are 2 m/s and 14 m/s. the phase difference between the waves at a point on screen having intensity less than the average intensity on screen may be (a) p 4 (b) 2p (c) p 3 (d) 7p 8 1 v 2 3v R Fig. Then find the ratio t1/t2. If each differes in phase from the next by p/4. dropping a vertical height h. Three simple harmonic motions in the same direction having each of amplitude a and the same period are superposed. Which of the following equations is/are not valid for this situation? C en ga (a) Resultant amplitude is ( 2 + 1) a (b) Phase of resultant motion relative to first is 90°. of the box at position 2. (c) The energy associated with the resulting motion is (3 + 2 2 ) times the energy associated with any single motion. A particle moves with uniform acceleration along a strainght line AB. diatomic and triatomic gases whose initial volume and pressure are same.0 m. then the triatomic gas will have maximum final volume (d) if the compression is adiabatic. Find the value of ‘*’. Monoatomic. M is the mid-point of AB. the characteristic X-rays spectrum will have .002% (c) 0.5 J K–1 and coefficient of linear expansion is 2 ¥ 10–5 °C–1.9 Paper 2 X ge E Le L (c) LE 2 2 R1R2 en ga LE (a) 2 R12 (a) Fnet (b) ma Pv t Frotor Fgravitational (c) Fdrag di a Frotor (d) Fdrag Frotor ma Fgravitational In Fgravitational 4.29. The total heat produced in R2 is . if an electron with an energy of 23 keV strikes a silver target. X is joined to Y for a long time. percentage increase in the volume of the plates is (a) 0.27.004% 2. B.L td . Assume 50% of their stored energy increases their temperature till the capacitor gets completely discharged and energy equally distributes over the plates. B.003% (d) 0. The plates are now connected by a thin wire of negligible heat capacity. Then the geometrical path length traversed by the light in the slab will be: (a) 2 3 (c) (b) 6 2 3 3 3 + 2 2 (d) 60° R1 m= 2m (3 / 2) 2 2m air m � (3/2) 2 LE (b) 2 R22 (d) Fdrag Fgravitational Y LE 2 R2 2 R12 3. B. Also the minimum energy of a striking electron so as to produce L X-rays is 3. B. In the circuit shown in Fig. A helicopter is moving to the right at a constant horizontal velocity.Appendix B: Mock Tests B. Which of the following diagrams can be correct free body diagram representing forces on the helicopter? C Frotor Z Fig.56 keV. If thermal capacity of each plate is 0. and then X is joined to Z.001% (b) 0.27 2 Fig. energy corresponding to Ka transition is 21. Fdrag and force on it caused by rotor Frotor .75 keV. For a silver atom. Now.29 5. A ray of light strikes a cubical slab as shown in Fig. A parallel plate capacitor of capacitance 200 mF is charged by a battery of emf 100 V.28 ar ni R2 Direction of motion ng 1. B. It experiences three forces Fgravitational . Section A: One or More Correct Answer Type 2 2m Fig. The battery is now disconnected and temperature of the plate is equal to atmospheric temperature. B. The heights of the tube above and below the water level in the vessel are h1 and h2.H. Ê 3ˆ (a) tan -1 Á ˜ Ë 4¯ Ê 3ˆ (b) cot -1 Á ˜ Ë 14 ¯ Ê 3ˆ (c) cot -1 Á ˜ Ë 4¯ Ê 3ˆ (d) tan -1 Á ˜ Ë 14 ¯ . 7. B and C. then 3Q 6Q (a) E = (b) E = 4 p e 0a 2 4 p e 0a 2 6. the gauge pressure of water in the highest level CD of the tube will be (a) 3. (a) vB = 2 gh2 Pv t ng 3 6Q 3Q (d) E = 4 p e 0a 4 p e 0a 2 8. Consider a body undergoing S.5 ¥ 104 N/m2 For Problems 12–14 A charged ball of mass 9 kg is suspended from a string in a uniform electric field E = (3i + 5 j ) ¥ 105 N/C. Give your answer in terms of angle made by string with vertical. Determine the velocity vB of the water issuing out at B. (a) F ◊ r (b) a ◊ r (c) v ◊ r (d) a ◊ F (c) V = A tube of uniform cross-section is used to siphon water from a vessel V as shown in Fig.10 Physics (b) vB = 2 g(h1 + h2 ) (c) vB = 2 g(h2 . B. find the new equilibrium position of the ball.1) W (a) The combined resistance in series is (8W ± 5%) W. Which of the following be always negative (throughout the motion). Take g = 10 m/s2.9 ¥ 104 N/m2 (d) 1.5%) W. The ball ge For Problems 9–11 ga C D h1 P0 C en A x q Fig.0 m. B. the net force acting on the body be F.3) and (5 ± 0. (a) 50 N (b) 100 N (c) 150 N (d) 200 N If the direction of electric field is reversed.L td . (a) 10 mC (b) 100 mC (c) 1 mC (d) 200 mC Find the tension in the string. the maximum value of h1.8 m 11.30 y q Le ar ni is in equilibrium with q = 37°. Find the charge on the ball. the acceleration of the body be a and its velocity be v.0 m (c) 7. will be (a) 3.30. If h2 = 3.0 m (b) 6. Section B: Comprehension Type P0 . 9. at any in stant of time. Three equal point charges (Q) are kept at the three corners of an equilateral triangle ABC of side a. h2 V B Fig. (b) The combined resistance in series is (8W ± 12%) W. In (a) Only Ka and L X-rays (b) Only L X-rays (c) K and L X-rays (d) Only Ka X-rays Two resistances are expressed as (3 ± 0. for which the siphon will work.0 ¥ 104 N/m2 (b) 5. 13. P is a point having equal distance a from A. If E is the magnitude of electric field and V is the potential at point P. the displacement of the body be r .31 12.h1 ) (d) vB = gh2 di a 10.9 ¥ 104 N/m2 (c) 3. The pressure over the open end of water in the vessel is atmospheric pressure (P0). (c) The combined resistance in parallel is (15/8 + 7. Given h1 = h2 = 3.2 m (d) 4.0 m. 14. B.M. Let. (d) The combined resistance in parallel is (15/8 ± 17%) W. A battery of emf E is connected across a conductor as shown in Fig. The disc takes time T to go from A to B. For Problems 15–17 In a Young’s double slit experiment set up.34 a 2m Fig. where y is in mm and t is in seconds. a slab of thickness 2 ¥ 10–3 mm and the refractive index 1. Between A and B. At t = 2 s. B. A constant horizontal force starts acting at the center. there is sufficient friction to prevent slipping. A disc of radius R is at rest at a point A on a horizontal surface.11 (r) Distance covered in (c) is less than that time T during part AB (s) Angle rotated by (d) is more than that disc in time T during part AB 19. Column II (a) q2 64pe 0 R (b) 5q 2 16pe 0 R (c) q2 16pe 0 R (d) None of these . Each of the two concentric conducting spherical shells is given a charge q. Match the columns. fringe width is b1. the source S of wavelength 6000 Å illuminates two slits S1 and S2 which act as two coherent sources. At t = 1 s. the position of central maxima is (a) 2 mm above C (b) 2 mm below C (c) 4 mm above C (d) 4 mm below C 17. Match the following: y S2 Pv t 1m x C 2 mm r2 E Fig.Appendix B: Mock Tests B.33 Column I Column II (p) Angular acceleration (a) zero of the disc (q) Angular velocity of (b) constant but not the disc zero q 2R Fig.L td . (s) If both are earthed. Then (a) b1 > b2 (b) b2 > b1 (c) b2 > b1 (d) Data is insufficient. B. but there is no friction to the right of B. B.34.5 is placed just in front of S1.32 ar ni ng 15. P is taken to be the origin of the co-ordinate system.35 Column I (p) The total energy of the system if none of the spheres is earthed. As one observer from A to B. At t = 0. The source S oscillates about its shown position according to the equation y = 1 + cos pt. R Le q Section C: Matching Column Type C en ga ge 18. B. B. then the total energy of the system. (r) Now original charges are restored and the inner sphere is earthed. To the right of B: F A B Fig. (q) The total energy of the system if the outer sphere is earthed. di P r1 S1 1 mm B A In S . fringe width of the figure is b2. The central maxima is formed at (a) 1 mm above C (b) 1 mm below C (c) 2 mm above C (d) 2 mm below C Column I Column II (p) Current (a) increases (q) Drift velocity of (b) decreases electron (r) Electric field (c) remains same (s) Potential drop across (d) cannot be deter the length mined 20. and at t = 2 s. 16. 38.1 J (d) –24 J 4. A current-carrying wire is placed in the grooves of an insulating semi circular disc of radius R. B. B.37. Then the charge on inner sphere is .3 J (b) –8. B. The current enters at point A and leaves from point B. Ram.38 (a) –4Q (b) –10Q (c) Zero (d) None 6.0 m/s. B. as shown in Fig. are moving with equal velocity. A conducting sphere of radius R and a concentric thick spherical shell of inner radius 2R and outer radius 3R is shown in Fig.36 moving up so that the block of mass M exerts a force 7Mg/4 on the floor of the box? a Section A: Single Correct Answer Type (a) Cart carrying Ram will have more speed finally than that carrying Shyam. A charge +10Q is given to the shell and inner sphere is earthed.B.0 m C i 30° B Fig. At some time snow begins to drop uniformly. How much work is done on the mass between A and B by the force of friction? (Take g = 9. (c) Cart carrying Ram will have same speed finally that of carrying Shyam. Pv t Paper 1 R ge V Le ar ni (a) g/4 (b) g/2 (c) 3g/4 (d) 4g 3.39 (a) m 0i 8p R 3 (c) 3 m 0i 4p R (b) m 0i 4p R 3 (d) none of these . B. B. (d) Depends on the amount of snow thrown. A 1-kg mass is projected down a rough circular track (radius = 2.39. B. picks the snow from cart and throws off the falling snow sideways and in the second cart Shyam is asleep. A B en ga 2. on which sit two twins of same mass.37 (a) –7. Two identical carts constrained to move on a straight line. 5. Determine the magnetic field at point D. it is 6. A i 30° C D R Fig. B. (b) Cart carrying Ram will have less speed finally than that carrying Shyam. sitting on one of the carts. After the snow fall is stopped.12 Physics Mock Test 3 In a ng M Fig. A monkey is climbing up a tree at a speed of 3 m/s.0 m) as shown in Fig.36 di 1.1 J (c) –6.L td . A dog runs towards the tree with a speed of 4 m/s. The speed of the mass at point A is 3 m/s and at point B. What is the relative speed of the dog as seen by the monkey? (a) >7 m/s (b) Between 5 m/s and 7 m/s (c) 5 m/s (d) <5 m/s 2.8 m/s2) 90° Fig. With what acceleration a sho-uld the box in Fig. The gravitational potential in a region is given by V = 20 (x + y) J/kg. (a) 40 2 N (b) 40 N (c) Zero (d) 20 N 10. the velocity is negative and acceleration is positive. the velocity is positive and acceleration is negative. What possible relation holds between magnitudes of normal reaction and friction forces. fB and fC are frictional forces and NA. Switch S is closed at t = 0 in the circuit shown in Fig. Which of the following statements is/are true with respect to the motion? Pv t 500 mH S where R is gas constant) form an ideal gas mixture. (c) Inclination of the string with vertical is p/2 – tan–1 (a/g) towards left.43 shows three blocks on a rough surface under the influence of a force P of same magnitude in all the three cases. C en ga ge P q q P (a) (b) P (c) Fig. The change in flux in the inductor (L = 500 mH) from t = 0 to an instant when it reaches steady state is 5W A di O In (a) 2 Wb (b) 1. fA. Figure B. which one is correct? a Fig. (b) Inclination of the string with vertical is tan–1 (a/g) towards right. Le ar ni ng Fig. (c) Between B and C. A cork suspended from the bottom of a container filled with water with a string as shown in Fig. the acceleration is positive.42 gives its displacement as a function of time. 12.13 20 V 5W 5W Section B: Multiple Correct Answer Type 50 mF t=0 11. B. both velocity and acceleration are positive. The molar specific heat at constant pressure of the mixture is (neglecting vibrational degrees of freedom) 7 16 10 23 (a) R (b) R (c) R (d) R 3 7 3 7 . The coefficient of friction is same between each block and ground. 9. B.42 (a) In the motion between O and A.L td .40. Two moles of helium. four moles of hydrogen and one mole of water vapor (for water vapors CV = 3R. NB and NC are reactions. (d) Inclination of the string with vertical is p/2 – tan–1 (a/g) towards right.41 (a) Inclination of string with vertical is tan–1 (a/g) towards left.43 . (Assume that blocks do not overturn about edge.41. 7. B. B. B. B. (d) Between C and D.5 Wb (c) 0 Wb (d) none 8. B.Appendix B: Mock Tests B. A particle has a rectilinear motion and Fig.40 B D C Fig. (b) Between A and B.) Here. If the container accelerates in a horizontal direction towards right. Find the magnitude of gravitational force on a particle of mass 2 kg placed at origin. Potential difference appearing across 6-W resistance is VA – VB = 10 V.25y cm. 14. B. two wavelengths of light are used simultaneously where l2 = 2l1. B.00-W resistor is 5 A. Calculate x.00 W ge 3. the position of image of a bird is at v = 1. (b) The center of mass of particles will move along a line inclined at some angle with vertical. In Young’s double slit experiment. Some currents are shown.00 A 3. In the figure a conducting rod AD makes contact with the metal rails BA and CD which are 50 cm apart in a uniform magnetic field of induction 0. In the circuit shown in Fig. Given q2 > q1 and v1 cos q1 = v2 cos q2.44.02 W. The tank is 20 cm deep and if a bird is flying over the tank at a height 60 cm above the surface of water. each link having resistance r is fitted in a resistance-free conducting circular wire.5 Wbm-2 perpendicular to the plane of the diagram. (a) maxima of wavelength l2 can coincide with minima of wavelength l1. ar ni 2. (b) fringe width of l2 will be double that of fringe width of l1 and nth order maxima of l2 will coincide with 2nth order maxima of l1. D C Fig. In the fringe pattern observed on the screen. B. B.46 ng (a) The center of mass of particles will move along a vertical line.00 W 3. Calculate y. In a finite square grid. (d) The particle (2) will be above the center of mass level when both particles are in air.00 W B Fig. The radius of curvature of mirror is 40 cm and refractive index for water m = 4/3. Pv t (a) NA > NC > NB (b) fA > fC > fB (c) fC > fA = fB (d) NC > NA = NB 13.45.45 (a) The current in the 4. The total resistance of the circuit ABCD is 0. 18. a B A 6.47 Work done in 10 seconds is found to be 100x J. B. (d) The resistance R is equal to 9 W.44 R E2 Le E1 4. B. . Two particles of equal mass m are projected from the ground with speeds v1 and v2 at angles q1 and q2 as shown in Fig. v2 v1 Section C: Integer Answer Type 16.14 Physics q1 (1) q2 m m (2) (c) nth order minima of l2 will coincide with 2nth order minima of l1 (d) nth order minima of l2 will coincide with (2n – 1)th order maxima of l1 . A concave mirror with its optic axis vertical and mirror facing upward is placed at the bottom of the water tank.00 W A di In A 17. Which statement/s is/are correct? Fig.L td . (c) The unknown emf E2 is 54 V.B.00 W 4.00 A B Fig. Calculate x if the equivalent resistance between A and B is 21r/8x. C en ga 15. (b) The unknown emf E1 is 36 V. (c) The particle (1) will be above the center of mass level when both particles are in air. E1 and E2 are two ideal sources of unknown emfs. N Fig. The wave pulse from the source S at t = 0 reaches the observer at time [for v1 < v2 < v3] ng Section A: One or More Correct Answer Type b a K di Paper 2 B O A In 20. empty box in which a block of mass m is hanging in equilibrium with the help of a vertical spring of stiffness K.5 m down the slope. B.15 19.48 a Pv t Determine the velocity (in m/s) of the axle of the wheel A. . B. Suppose there is an. B.50. The system is released from rest. Now the box is moved downwards in water (rw = density of water) with a constant acceleration a by applying a vertically downward force F as shown in Fig. when it has travelled a distance s = 3. Consider two cases: (i) the rays parallel to the principal axis are incident on plane surface (ii) the rays parallel to the principal axis are incident on curved surface Which of the following is correct? (a) The rays in both the cases will converge at same distance from the surface on which rays are incident.cos 2q ) .49.50 F K V2 (a) t = L v2 (b) t = L v1 (c) t > L v3 (d) t < L v3 3.49 (a) 2p K (F ) M ( rwVg ) (b) 2p K ( rwVg ) M (F ) C S Fig. which rotates about a fixed horizontal axle O. A plano convex lens has thickness of 4 cm and radius of curvature of its curved surface is 30 cm. passing over a pulley C. B. Both wheels and the pulley are assumed to be homogeneous disks of identical weight and radius. B. Light of intensity I falls along the axis on a perfectly reflecting right circular cone having semi-vertical angle q and base radius R. thus pulling up the wheel B which rolls along the inclined plane ON. as shown in Fig. The volume of the box is V. Neglect the weight of the string.48. [Take a = 53° and b = 37°] (c) 2p K ( F + Mg ) M ( rwVg + Mg ) (d) None of these 2.L td . An observer (O) and source (S) move horizontally with speeds v1 and v2 as shown in Fig. B. If E is the energy of one photon and c is the speed of light. The string does not slip over C. 2c . v3 is the velocity of sound wave. At t = 0. the horizontal separation between O and S is L. calculate b if the force acting on the cone due to light is found to be p R2I b (1 .Appendix B: Mock Tests B. C Le ar ni 1. The wheel A rolls down the inclined plane OK. The time period of oscillation of the block in the frame of the box is V1 O L ge ga M en Water Fig. A wheel A is connected to a second wheel B by means of inextensible string. If L1 = L2. then ng should be used. The image of the virtual point object O formed by the lens LL is (a) Virtual (b) Real ar ni R Le I . The frequency in L1 and L2 are same. (b) Some part (not 100 %) of the energy supplied by the battery will be dissipated in R and remaining will continue to store in L. At x = 0. (d) None of these.0 (b) 4. respectively. B. A standing wave is created and the “junction” is found to be a node.16 Physics (b) To calculate the focal length of lens.L td .51 C en ga (a) Voltage across R is RI. the potential difference used to accelerate the electrons is increased from 12.46 px – 92 pt) (All parameters are in MKS): 9. the difference between the wavelength of Ka-line and minimum wavelength increases two-fold. 4. t) = A cos (0. Section B: Comprehension Type For Problems 9–11 Two plane harmonic sound waves are expressed by the equations: y1(x.53. The % error in the volume of the cylinder is (a) 3. The wavelength of the Ka line is (a) 1 Å (b) 0.0 (d) 2. how many times the amplitude of y1 + y2 is zero in one second? (a) 192 (b) 48 (c) 100 (d) 96 . The minimum number of loops in L1 is 1. 2 7. (a) (b) (c) (d) L1 L2 m1 m2 Fig.5 Å (d) None of these 5.01 cm was used to measure diameter of a cylinder as 4 cm and a scale (0–15 cm) with the least count of 1 mm was used to measure a length of 5 cm.5 6. (c) Voltage across L is equal to zero. B.52 as paraxial. which of the following given statements is/are correct? di (c) To calculate the position of image (c) Located below the principal axis (d) Located left of the lens In L ge Supply voltage Fig. B. are joined and tied horizontally by two fixed vertical walls as shown in Fig.8 kV. B. How many times does an observer hear maximum intensity in one second? (a) 4 (b) 10 (c) 6 (d) 8 10. As a result.0 (c) 5. where L is a pure inductor. In a Coolidge tube. B. Consider the rays shown in Fig. The wavelength in L1 and L2 are different.5 px – 100 pt) y2(x. R R f 2˚ Î 1 L O (Principal axis) 1 1 1 = f u v F (First principal focus) L Pv t Fig. What is the speed of the sound? (a) 200 m/s (b) 180 m/s (c) 192 m/s (d) 96 m/s 11.53 The minimum number of loops in L1 is 2.B. 1 (d) Magnetic energy stored is LI 2 .4 kV to 24. When current (I) in R-L series circuit becomes constant.52 a 8. Two strings of lengths L1 and L2 and mass per unit length 4m and m. È1 1 1 ˘ = (m – 1) Í ˙ is applicable. A Vernier caliper with a least count of 0. t) = A cos (0.5 Å (c) 1. If however thin beam is focussed at A. (s) Tension along the (d) Independent of length of string. the elongation of the spring is (a) Zero ng –Q m k a (c) 2 k (b) ˆ 2 Ê Q2 .0 (d) 2. (a) 1. Then Column I Column II (p) Pulse speed v along (a) Constant the string at any time.7 (c) 2. then 12. B. the pulse speed is v and wavelength Fig.56 of the pulse is l. (q) Acceleration of (b) Variable pulse at any time along the string. 15.54 (a) 2 k Ê 2 Q2 ˆ (b) mg Á ˜ k Ae 0 ¯ Ë (c) 2 k Ê Q2 ˆ Á mg . the maximum elongation of the spring is Pv t For Problems 12–14 A parallel beam of light falls on a solid transparent sphere. (r) Frequency of the (c) Depends on wave pulse when it travels length along string.5 14.mg ˜ Á k Ë 2 Ae 0 ¯ Ê Q2 ˆ Á 2 mg . neglect dielectric effect of the spring and Q2 < 2mAe0g. At any time t.2 Ae ˜ (d) None of these Ë 0¯ (a) T > 2p (c) T = 2p d m Fig. wavelength . Here the speed of the pulse is v0. then find the refractive index of the sphere. The plates are joined by a nonconducting spring of force constant k. A uniform rope of mass m and length l hangs vertically from a rigid support.Ae ˜ (d) None of these Ë 0¯ Section C: Matching Column Type ar ni Le ge ga Wall en C Smooth table k Ê 2Q 2 ˆ mg Á Ae 0 ˜¯ Ë 16.17 Fig. di m k In 17. 18. Assume the pulley to be massless. If the block is released from rest. the initial separation between the plates. B. A transverse pulse of wave-length l0 is produced at the lower end. then the whole beam can be focussed at A. (b) The whole beam can be focussed at A only if the beam is thin enough (c) If the beam is thin.5 (b) 1. Which is correct? (a) If the beam is thick. (d) None of these 13.5 (d) none of these For Problems 15–17 The plates each of area A of a parallel plate capacitor are given charges Q and –Q. A Read the paragraph carefully and answer the following questions.55 (b) T < 2p m k (d) T will depend on Q. the thin beam can be focussed at centre of sphere. (a) 1. The natural length of the spring is d. If Q2 = 2mAe0g. The left plate is connected to a vertical wall through a massless non-conducting rope and the right plate is connected to a block of mass m through similar rope. Q . If T be the time period of oscillation.Appendix B: Mock Tests B. B. then the beam cannot be focussed before A. For what value of refractive index µ.5 (b) 2 (c) 2.L td . Ans (d): Displacement y (x. Column I Column II (p) Magnitude of flux (a) pR2E through base of cone Section A (b) pR2E/2 Magnitude of flux through curved part of cone Magnitude of flux through curved part MNQP of cone Net flux through entire cone Pv t P (q) (r) (s) .58 ng Le 1. Q = sT4. DQ is the heat supplied to gas and DW is the work done by the gas. Its base radius is R. It is kept in a uniform electric field E parallel to its axis. t) = A cos (kx – wt) At t = 0. Ans (a): N – mg cos 60° = ma cos 60° N = m cos 60° (a + g) = 400 N a 60° (d) non-zero M ar ni Paper 1 a cos 60° (c) Zero h/4 Column I Column II (p) During AB (a) DQ > 0 (q) During BC (b) DQ < 0 (r) During CD (c) DW > 0 (s) During DA (d) DW < 0 20. the center of mass will move in forward direction and the net torque about the center of mass is in anticlockwise sense. 19. 3.25 ¥ s ¥ (2T)4 = 0. Q¢ = esT ¢4 When T ¢ = 2T and e = 0. B.B.57 P ge ga N 60° Fig. B. A cyclic process for an ideal gas is shown in Fig. S.58 is vertical.2 vy vy ay x ay 6 5 vy = 0 ay .25 ¥ 16 ¥ sT4 = 4sT4 = 4Q 4.57. The axis of a hollow cone shown in Fig. point 1 on the string has maximum displacement Velocity vy (x. Match the columns.25. Hence.L td . Q h a di In E N R h/4 E Fig.18 Physics B C A D T Fig. Ans (b): The spool is having the tendency to rotate in clockwise direction. Q¢ = 0. Ans (b): By Stefan’s law. Solutions of Mock Test 1 Rate of dissipation of heat by a body of emissivity e. The point of contact of spool with ground will move in backward direction. B. S. friction will act in forward direction. B. t) = wA sin (kx – wt) vy 1 9 vy 2 8 ay ay 3 4 vy = 0 vy ay = 0 7 ay Fig. 2. the rate of dissipation of heat by a black body.1 C en Reading will be 40 kg. e.v f = 20 i . pl l x = = 2 ¥ 2p 4 i. Ans (d): According to the problem. 20 ng At t = 0.Appendix B: Mock Tests B.5 cm . Ans (c): Here B (due to I) and dl will be in same direction.3 ga 7. Hence. Hence forces (II > I). Then to go from O to A or from O to B.20 j a= a 2 3 2 p a2 - a2 2 . vp = w a 2 .4 can be redrawn as: w= Now. i. Ans (c): A1 = 2. Let us assume that particle moves from P to A and back from A to P in 0.4 2 2W Le 2W 2W 2W Fig. ay is the maximum downward acceleration.e.5 s..5 cm2 v2 In di 10.5 s. time taken 1 is s. 1/4 A P O B (Mean position) Fig. t) = – w2A cos (kx – wt) At t = 0. (b) is correct. i. (c) is correct. the particle is at P when t = 0. force will be towards right. B ◊ dl = B dl cos 0∞ en Ú C Ú = B dl Ú = B dl = BN 2p r = N(B2pr) = Nm I Ú A1v1 = 1. the string has maximum upward acceleration. 2 4 4 We have x = a sin wt. (a) is correct. 5. ay = (x. (d) is correct. S. vy is maximum at kx = p/2. where O is the mean position. Velocity vy is positive and acceleration ay is negative when 0 < kx < p/2. Ans (c): In (I) and (II). Hence. Ans (a): Fig.x 2 = p 0 8. but in case (II) distances are minimum. time taken for the particle 2 1 to go from P to A is s. S.. vy and ay are positive when p/2 < kx < p. vy and ay are negative when 3p/2 < kx < p. flag: vw / f = vw . vy is negative and ay is positive p < kx < 3p/2.19 VW / F -20 Fig. Hence time taken to go 4 1 1 1 from O to P is tO Æ P = . In case (III) force on P will be towards left and in (IV).r.. forces on P due to all three charges are in same directions.5 s. At t = 0. So.5 + 1. point 9 on the string has maximum downward acceleration. Acceleration. But on calculation forces (IV > III).5 This implies that T = 0.5 s and once again at P when t = 0. If cos kx = 1 fi kx = 2 p fi x = l. Ans (c): v f = 20 j .t. 3 = p Velocity of wind w.5 = 2 s 2p =p T Let OP = x. As per the problem. point 3 on the string.L td . v1 = 3 m/s. vw = 20 i i. x = l/2.8 m. h = 0..e. S.e.= s. B. g = 10 m/s2 v22 = v12 + 2gh = 25 fi v2 = 5 m/s ar ni A S E . Hence. 1ˆ a Ê When the particle is at P. Hence. 6. kx = –p. x = a cos Á p ¥ ˜ = Ë 4¯ 2 ge B 2W A1v1 = A2v2 fi A2 = a Pv t 9. from P to B and back from B to P in 1. L td . v2 = 7 m/s v1/cm = v1 – vcm = 4 – 5 = –1 m/s v2/cm = v2 – vcm = 7 – 5 = 2 m/s 14. Ans (a. Ans (3): The length of the string is 1. d): 1 . If work is done on the particle. 2 and 3. c): f1 : f2 : f3 = In vcm = ng Pv t Section C a 3 m/s 6 m/s m2 = 2 kg m1 = 4 kg m2 m1 Note that the transmitted wave is always in phase with the incident wave. PB – PA = P0 + rgy2 – (P0 + rgy1) = rg (y2 + y1) where y2 – y1 = AB sin q = b sin q Then. Therefore. either both towards right or both towards left. S.002 nm–1 lo2 1 = 0.B. d): The amplitude of the transmitted Ê 2 v2 ˆ wave is given by at = ai Á Ë v + v ˜¯ 3 2 ¥ p = 3 2 m/s p v1 v2 16. Ans (a. 13. d): If any work is done on the particle (work will be done by tangential forces and not by centripetal forces). \ 32 . reflected and transmitted waves are always equal.004 nm–1 lo3 fi lo3 = 250 nm = 2500 Å Hence.Q 18. The frequency of the incident. If V be the velocity of transverse waves in the string. 12. we have PB – PA = rgb sin q 17. c.20 m. C h1 y2 h1 q B q Fig. Ans (8): Let the power lost to surrounding is Q. Ans (4): Since time of contact is equal to time of half cycle. at may be more than ai. 2 Section B 1 = 0.7 Then.20 Physics Now. the fun- . velocity and acceleration both should be in same direction. b): For increasing speed. S. Ans (a. Otherwise. therefore T= p m p = k 4 19. di 11. b. at /ai < 1 only when v1/v2 < 1.001 nm -1 lo1 en A y2 – y1 ar ni u1/cm = u1 – vcm = 6 – 5 = 1 m/s u2/cm = u2 – vcm = 3 – 5 = 2 m/s Hence (b) is incorrect. slope tan q = e ga lo1 = 1000 nm = 10000 Å. After collision: v1 = 4 m/s.6 4¥6+2¥3 = 5 m/s 4+2 y1q Le hc hc hc 1 1 1 : : = : : = 1: 2 : 4 lo1 lo2 lo3 lo1 lo2 lo3 From Einstein’s photoelectric equation hc = f + eVs l fi Vs = ge hc f el e hc Therefore. vmax = aw = 15. then its speed will change. UV light can be used to eject photoelectrons from all the metals 1. Ê dm ˆ 16 – Q = Á S (10) Ë dt ˜¯ ÈÊ dm ˆ ˘ and 32 – Q = 3 ÍÁ ˜¯ S (10)˙ Ë dt Î ˚ 1 = 0. fi lo2 = 500 nm = 5000 Å.Q = 3 fi Q = 8 W 16 . Ans (a. it will go in the form of KE. Ans (6): The pressures at A and B are PA = P0 + rgy1 PB = P0 + rgy2 Fig. Ans (a. VP > VR. VS = VR. Ans (b): Angle of refraction = Angle of incidence + Angle of deviation fi r = 30° + 15° = 45° m1 sin 30° = m2 sin 45° fi m1 1/ 2 m 1 = = 2. ln > 30 cm The string vibrates in its third overtone. angle of incidence ª 90°.10 . Let the current in the circuit is I. then the displacement of 2m is x + 2L and displacement of m is x – 2L towards right. which have wavelengths 240 cm.21 P 15 cm N Q Fig. VQ > VR. N1N2 = 2p l p a ◊ = a cos = l 8 4 2 Hence. a = ( 2 ) (3. damental frequency is f = Dxcm = m1Dx1 + m2 Dx2 + m3 Dx3 m1 + m2 + m3 fi 0 = 3mx + 2m(x + 2L) + m(x – 2L) L fi x = 3 3.9 C en So VP = VQ. 80 cm and 60 cm. the string vibrates in one of the first four harmonics. From symmetry. S. m 3m 4L Fig. no current will flow in the loop.40/n.L td . 2 = m2 1/ 2 m1 2 When ray enters from medium (2) for maximum deviation. P and Q represent the space point of string where amplitude is 3.5 a cos kx = a cos a N a di 3.5) = 5 mm 3. i. the string vibrates in its fourth harmonic. Ans (c): Pav = EvIv cos f = 100 ¥ 10 cos f = 1000 cos f Watt So depending upon the value of f. VP > VS. S. 120 cm. S. Ans (d): Since the entire loop is in magnetic field. Q Le P e ge e ga S R Fig.2= 0 ar ni Section A x 8¥5 = 25 A 8 ¥ 0. It is required to find a. Also if the equation to the loop is y = a cos kx with x = 0 at the antinode A. 4. 2.5 mm A Y = 3. Figure S. ng Since PQ = 15 cm. maximum deviation = 45° 5.2 In Hence.5 mm Hence. the wave length of this harmonic is ln = 2..5 mm with PQ = 15 cm. the amplitude at the antinode. Ans (b): hn1 = hn 0 + hn 2 = hn 0 + 1 mv12 m 2 1 mv2 2 . Hence. we see that N1P = PA = AQ = QN2 = 7. shows one of the loops and the data given in the problem. Ans (0): There are eight batteries. so no net emf will be induced in the loop. Pav can be equal to or less than 1000 W.8. we see that Pv t V . 20. PD across voltmeter = potential difference across rightmost battery = E – Ir = 5 – 25 ¥ 0.e.8 Then I = ln > 15 cm 2 Paper 2 1. ln = 60 cm. respectively. Hence. If it vibrates in the nth harmonic. The ends of the 2 ¥ 1.5 cm so that N1N2 = 30 cm = ln /2. as the string will vibrate with n loops.Appendix B: Mock Tests B.20 string are nodes and the string may vibrate in any of its higher harmonics. Ans (d): Let the box moves towards right by a distance x. from the figure. \ m2 sin 90 = m1 sin q fi q = 45° Therefore. L td . fi v12 . C en ga ge 9.˙ 2p r dr = 2p J 0 Í ˙ 0 Î R˚ ÍÎ 2 3R ˙˚ Apply ampere’s law: B2pr = m0I fi I = ng Lc l Section B Ú È 3R / 4 (3R / 4) 2 ˘ 3R Bmax = m0 J 0 Í m0 J 0 ˙= 3R ˙˚ 16 ÍÎ 2 15.12 2 Fig.m g cos q = a 2 2 2 Solve to get T = Mg sin q/8 Ans (b) Ans (b) For A: Pv t fi h (n1 . always upwards and hence constant.11 A 0 Ê r 2 r3 ˆ È r r2 ˘ B 2p r = m0 2p J 0 Á fi B = m0 J 0 Í ˙ ˜ Ë 2 3R ¯ ÍÎ 2 3R ˙˚ dB To find max B: = 0 fi r = 3R / 4 dr ar ni C C a T /2 M T B b r È r 2 r3 ˘ r˘ È J 0 Í1 . as l keeps changing direction. Ans (d): Current within radius r: I = In Lo .22 Physics 1 m (v12 . c. Its vertical component also remains same.n 2 ) = t dt t Fig. let it be b.13 dA = Vdt Ú dA = Ú ÁË L dt ˜¯ dt A= Ê dI ˆ dI ˆ Ê ÁËQ V = L dt ˜¯ I2 Ú L dI = L (I 2 . 13.n 2 ) m Ans (b. c): According to Lenz’s law. Ú r a di dA Le LC = rmv. S. when switch is opened the magnetic flux linked with coil B decreases. Ans (a): Given I1 = 0 V M T M q 0 Fig. d): Concept based. c.T = a 2 2 M M M For C: T + g sin q .14 4 t .v22 ) 2 6. S. Ans (c): Area under V–t graph. so L0 also changes in direction but its magnitude remains constant. Therefore. Ans (c) 10. Similarly. S. Ans (b.v22 = 8. Ans (a. The direction of induced current in coil B will be clockwise. Ans (d): Here the acceleration of A and C will be same. Ans (a) 11. S. current in coil A grows in clockwise direction (when switch S is closed).e.I1 ) I1 16. current is induced in coil B in anticlockwise direction.. O m 14. i. always perpendicular to r. 7. The acceleration of B will be independent of A and C. 2h (n1 . This causes increase in magnetic flux linking coil B in clockwise direction. For B: T + Mg sin q – T = Mb fi b = g sin q a T /2 J dA V Fig. d): Angular momentum about O: L0 = lmv. 12. M M g sin q .B. let it be a. (p – b) (q – a) (r – d) (s – c) Theory based.R) 2 1+ R È r . 19. d) (r – d) (s – c. If E decreases. hence df 0. Ans (b): If a body is projected from the surface of the earth with a velocity v and reaches a height h.L td . so E also decreases. A dR = a I2 = dx Pv t 0 . As area increases. I = neAvd. (p – c) (q – b) (r – b) (s – b) Current through any cross-section should remain same as all the cross-sections are in series.1 0. Ans (a): f = uv = uv (u + v) -1 u+v È u ˘ df = v Í(u + v) -1 ˙ du (u + v) 2 ˚ Î È v ˘ + u Í(u + v) -1 ˙ dv ( u + v) 2 ˚ Î df du du dv dv = + f u (u + v) v (u + v) Since all errors add up with positive sign.2 = + + + f 40 60 20 60 0. (p – c) (q – b.3 0. r dx . S.Appendix B: Mock Tests B. It can also be done like this dR 12 I A1 2 x 3 4 t Fig. 17.1 0.1 0.75% . en Here v = kve = k 2 gR C 1 mg (r . then according to the law of conservation of energy.15 Fig.23 V E is directly proportional to vd. 20.2 0. A increases with x.R or k2r = r – R R ˙˚ Î or r = R 1. S.R ) mk 2 .7 ˜¯ 100 = 40 ¥ 100 = 1.R˘ k 2 R Í1 + = r . a) W is positive during expansion and negative during compression. so drift velocity decreases. Ans (d): A1 = 21 Le Paper 1 Section A 1 mgh mv 2 = 2 1 + h/ R ga ge 1.16 A1 21 = =7A 3 L di Section C Potential across dx: dV = I dR = I r dx A dV I r 1 = µ dx A A In E= So E and dV/dx both decrease as A increases. I2 is maximum if the area is maximum. ar ni ng 18. then PD across a segment of fixed length also decreases.3 gR = (r .7 + = 40 20 40 = Ê df or Á Ë f ˆ 0. Maximum area will be at 4 s.k2 Solutions of Mock Test 2 2. V≤ = V¢ + 2(2V) = 5V + 4V = 9V 4. peak voltage E0 = 60 V 6.v1 u1 .DU = a ln h .v1 ) fi 1 = 0 u1 . Ans (b): Final charge on capacitor qf = CE Charge flown through battery: Pv t fi v1 = u1 + 2v0 DU = CVDT = a .1˚ 9. upthrust = 0. Ans (a): E = 120 sin 100 pt cos 100 pt = 60 sin (200 pt) So.B. because there is no dissipating force (such as friction) which decreases the energy of system. hence torque of F3 is greatest. Ans (a): Upthrust = Weight of fluid displaced. d): Pi = P Pf = 2P Vi = V Vf = ? .1) T0 r -1 di v2 .17 .1˘ W = Q .v .0 ˜ + Heat Ë 2C 2C ¯ ge CE 2 Wb = DU + Heat q = q f . V¢ = V + 2(2V) = 5V Velocity after second collision. b. 10. O D R 2 a A E Fig.j + k ] hT0 After collision 8.24 Physics 3.( . the acceleration of block will be changing both in magnitude and direction. 12. c): Here. We can apply the conservation of energy. 5. S. b. S.L td . so we cannot apply the kinematical formulae of constant acceleration. Ans (c): Use the following result: v0 v0 O u1 v1 Before collision l = AC = 2 R (cos 45∞i + sin 45∞ j ) = 2 (i + j ) F = Il ¥ B = 1 ¥ 2 (i + j ) ¥ (3i + 4 j + k ) = 2 [i . Ans (c): a = 60° . Ans (a.19 Solid angle subtended by ABDE: w(ABCDE) – w(BCD) = 2p – p = p q p q Hence.(. Ans (b): Force on semi-circular wire will be same as the force on straight wire AC.v0 ) R (h . As g = 0. Ans (a. Ans (c): Perpendicular distance of F3 is greatest from O. Solid angle subtended by BCD: w = 2p (1 – cos a) = p ar ni Ê q2 q2 ˆ qE = Á . Ans (b): Q = È h .q0 = Le 1 CE 2 8 ga fi Heat = C en 7. 2 Fig. S. y C A 2.Í ˙ RT0 Î r .18 R=1m x C B ng Using the above result: Now velocity after first collision.u2 = a ln h 0 In e= hT0 Ú CdT = a ln T T0 Fig. flux through ABDE: f = = e 0 4p 4e 0 Section B 11. But in options (b) and (c). ga ge 8+x 2–x –8 en 16. d): A = a + 2a cos 45° = ( 2 + 1) a Energy is proportional to the square of amplitude Maximum velocity is proportional to amplitude. S. Hence. Ans (2): Let the required distance is x.23 5 c . there will be no tension in the rod. Ans (2): Let m = Moon’s mass. roll will be under tension. Ans (a.L td . Ans (b. x = R/2. S. Ans (2): v 2 = 22 + Fig. c): In options (a) and (d). 8+ x 2. S. we get v2 – 4 = 196 – v2 fi v = 10 m/s v .2 10 . S.v 14 . d): Let maximum intensity be I0 I Df I = I0 wt Iav = 0 2 2 1 Ê Df ˆ fi cos Á < Ë 2 ˜¯ 2 18. vf is minimum Ë 2¯ For isothermal process. Hence. a p/4 17.21 2 2as 2 2 m/s 2 x Section C fi v – 4 = as t1 1 = t2 2 2 Fms M S rme = 2 Fme M e rms Putting the values. fi 196 – v2 = as From (1) and (2).Appendix B: Mock Tests B.v 2 = 2as 2 fi –6 = 2x fi x = –3m fi t = y 14 m/s Fig. a A fi di p/4 fi 14 . length will increase.x 19. Ms = Sun’s mass Me = Earth’s mass rms = the center-to-center distance from the Sun to the Moon rme = the center-to-center distance from the Earth to the Moon Fms = gravitational force exerted on the moon by the Sun Fme = gravitational force exerted on the moon by the Earth In Then Fig. Then. w = v/x = 3v/(2R – x). Ans (4): t = = c c Le p 3p < Df < 2 2 15.2 ¥ 105 Í = =2 ˙ ˜ Á ˜ Fme Ë M e ¯ Ë rms ¯ 16 Î 400 ˚ ar ni Ê Df ˆ 1 < I < Iav fi cos2 Á Ë 2 ˜¯ 2 ng 14. PV = (2P)vf fi vf = v/2 13.25 For adiabatic process PV3 = (2P) v3f 1/ 3 Ê 1ˆ fi vf = Á ˜ for v is maximum.2 8 Now t1 = = = a a a (1) �3 Fig. c. Ans (b.10 4 = = a a a a a (3) Pv t A t2 = (2) .22 C v S/2 t1 M S/2 t2 B 142 . c.20 2 2 Fms Ê M S ˆ Ê rme ˆ 32 È 1 ˘ =Á = 3. 24(b).6 = fi Q = 20 ¥ 10–6 C 2 10 There is no force ma. 5 32 + y 2 fi y = 4 Q2 20. 6. ma is the resultant of all the forces acting on a body.24 Wings of helicopter will push the air perpendicular to their plane of rotation as shown in Fig. the energy of Ka transition) = 21.B. S.1 ˆ Rse = (3 + 5) ± Á ¥ 100˜ = (8 ± 5%) W ¯ Ë 3+5 (a) (b) Fig. Ans (b): L = 2 m + 2 m + 2 m = 6 m Section A 1.L td . Ê r ¥ r ˆ Ê Dr Dr Dr r ˆ RP = Á 1 2 ˜ ± Á 1 + 2 + 1 2 ˜ ¥ 100 r2 r1 + r2 ¯ Ë r1 + r2 ¯ Ë r1 Ê 5 ¥ 3 ˆ Ê 0.5 ºC Now . Ans (c): 2 ¥ S ¥ DT = .e. Ans (a): When X is joined to Y. Ê 0.05) ¥ 100 = Á ± 17%˜ W Ë 8 ¯ 8 .003 % 2. 3. Moreover. Ans (a.3 + 0.24(a).1ˆ =Á ± + + ¥ 100 5 3 + 5 ˜¯ Ë 5 + 3 ˜¯ ÁË 3 = 15 Ê 15 ˆ ± (0. Basically. d): In series. Now let us see how Frotor acts. Ans (b): Energy required to remove an electron from K shell = energy required to remove an electron from L shell + Energy difference of K and L shells (i. hence only L X-rays will be produced.. ¯ 2 ÁË 2 where S = 0.1+ 0.3 + 0.1 0. here a is zero because the velocity of helicopter is constant.02 + 0. Fdrag acts opposite to the direction of motion.56 = 25.26 Physics From (1) and (2). S.31 keV The incident electron does not have this energy.75 + 3. 4. current in L:I = E/R1 In DV = g (DT) = 3 a (DT) V Energy stored in L: U = Q 20 = mF = 2 mF 10 10 Pv t 1 Ê1 ˆ CV 2 ˜ . percentage change = 3 a (DT) ¥ 100 % = 0. Q 100 ¥ 10. C en ga Frotor 2m 45° 2m ar ni 1 2 E2L LI = 2 2 R12 45° 60° ng Therefore. Ans (c): Fgravitational acts downwards. air will apply force Frotor on the wings as shown in Fig.3 0. Ans (2): = 100 2C (1) Q = 16 C (2) Paper 2 \ C = 60° a di ge Le This whole energy will be dissipated as heat after X is connected to Z. S.5 J K–1 (for each plate) \ DT = 0. TIR 45° 45° 2m TIR Fig. S. From third law.25 5. pressure h2 \ Enet 1 Q r 4pe 0 r 3 = E1 + E2 + E3 = Q Ê 3 ˆ 6Q = a˜ k = k Á3 4pe 0 a 3 Ë 2 ¯ 4pe 0 a 2 Pv t V B 1 Q ( P0 + Q0 + R0 ) 4pe 0 a 3 P0 a Fig. (labeled by subscript 1) 1 .27 7.h2 = . F is negative and vice-versa.2 m – 3. where h2 is the difference in levels A and B. 0).. Ans (a. Ê ˆ 3 P ∫ (–a/2. 12. a Mean position ge F r ga h1 P0 ˆ a˜ . S. we get P1 = Pressure at level CD = Patm – (rg)(3 + 3) = 1. . b): Refer Fig. 0) and R ∫ Á 0.0 m = 7. In y R 2 1 È r 2 gh2 ˘˚ .0 ¥ 105 ..(3) The minimum value of P1 = 0 (P1 cannot be negative because then not water will reach the level). S.8 N/m2 = 5.27 en When r is positive.26 8.L td . 0.9 ¥ 104 N/m2. Q ∫ (a/2. Ans (a) 14. S. CD. S. Hence putting P1 = 0. x P1 = PA - ar ni P ng O Q Le Fig. 0˜ 2 Ë ¯ (h1) max = . Ans (c): We apply Bernoulli’s theorem for level A and for the highest level.8 = 10.28 di 10. S. a.. v1 = vB = 2 gh2 . Ans (b) 13. Ans (d) T sin q = 3q ¥ 105 (1) T cos q = mg – 5q ¥ 105 (2) Solve to get: q = 100 mC T = 50 N After the reversal of the direction of electric field T¢ sin a = 3q ¥ 105 ⇒ T¢ cos a = mg + 5q ¥ 105 a F r Fig..Appendix B: Mock Tests B. we have E = O Patm 1.r gh1 2 Î = Patm – rg (h1 + h2) . Hence.0 rg 1 ¥ 103 ¥ 9.(2) From (1) and (2). 0. Section B C 9. ¯ Also.27. coordinate of point at which E has to be Ê 2 a calcualted is 0 ∫ Á 0. 3 Ë 2 3 A P0 = PA atm..28) and the points A and B are open to the atmosphere.0 m in equation (3) above. Ans (a): As long as water fills the tube (as shown in Fig. D . the velocity at B will be given by Torricelli’s theorem. vB = 2 gh2 .2 m 11. Ans (b.(1) to get PA = P1 + rv12 + r gh1 2 Since the tube has uniform cross section and water is incompressible.0 ¥ 105 – 6rg The gauge pressure at level CD = 6rg = 6 ¥ 103 ¥ 9. Ans (b): Putting h1 = h2 = 3.3. c): Let the points charges have the coordinate C P0 Therefore.. so the torque about center becomes zero. 16.L td . then PD across a segment of fixed length also decreases.B. Ans (d): Path difference between the waves reaching A.28 Physics 5 q (3 ¥ 10 ) q 17. then dy dy1 Dx = + . Also angle rotated is also more.1) t1 d Pv t T cos q 5 q (5 ¥ 10 ) T D1 = 1 m S2 2R D2 = 2 m Fig. Ans (c): Fringe width will be independent of the position of S.30 ga ge 15.1) t1 D1 D2 At t = 1 s. so E also decreases. So. To the right of B.R C2 = 4pe0 (R) = 8pe0R In Ê 3ˆ fi a = tan -1 Á ˜ Ë 14 ¯ a 2 ¥ (1. S. (p – c) (q – b) (r – b) (s – b) Current through any cross-section should remain same. S A C2 S1 y1 y C1 R C en 18.5 . y = 0 and for central maxima. as area increases so drift velocity decreases. S. I = neAvd. mg Fig. S.1) ¥ 2 ¥ 10-6 = 10-3 m = 1 mm 2 ¥ 10-3 a tan a = di Dx = . S. angular acceleration becomes zero. d) (r – d) (s – d) To the right of B.( m . Dx = 0. (p – b) (q – c) (r – d) (s – d) 4pe 0 (2 R ¥ R ) (p) C1 = = 8pe 0 R 2R . (p – a. If E decreases. q T sin q dy1 dy + D2 D1 For central maxima: Dx = 0 fi y1 = –2y = –2(1 + cos pt) At t = 2 s.32 q -q 2q . 19. c) (q – b. there is no friction. 20.29 = 3q ¥ 105 mg + 5q ¥ 105 = 3 ¥ 10-4 ¥ 105 3 = -4 5 14 9 ¥ 10 + 5 ¥ 10 ¥ 10 Section C T ¢ cos a ng a T¢ ar ni 3q ¥ 10 T¢ sin a 5 5 Le 5q ¥ 10 mg Fig. E is directly proportional to vd. angular velocity becomes constant. y1 = –4 mm Ans (a): If slab is placed in front of S1.31 Fig. D fi y1 = 2 ( m . it will cover more distance in same time T because now there is no opposing friction. Hence. 1 = – 6. Ans (c): FBD of M: If M exerts force F = 7Mg/4 on floor. U = (r) PD of inner sphere is zero.L td . U = –X -q q C1 a Fig.34 x2 q2 (q + x)2 + = 2C1 2C2 32pe 0 R (s) The charge on both will become zero.Appendix B: Mock Tests B. Ans (c): VD1M = VD .33 In Section-A Le 1.29 q ˆ Êx V = k Á + = 0 fi x = – q/2 Ë R 2 R ˜¯ (q) 2q charge will flow in the earth. fi v¢ for Ram and Shyam will have equal velocities.Mg = Ma 4 3g fi a = 4 3.0 ¥ 9. where m¢ is the mass of snow that is added. Ans (c): mv0 = (m + m¢)v. di 2R X Pv t R R 2R q+x U = Fig. then from the third law.35 . Ans (a): Potential of sphere should be zero. ar ni Paper 1 ng Solutions of Mock Test 3 v¢ = mv0 m + m¢ Since Ram throws off show sideways.VM = 4i .6 J Total KE at B if there had been no friction C KE at A = = 4. 5.6 J Gain in KE from A to B if there had been no friction = 19. Ans (c): Conservation of energy principle 1 ¥ 1.02 = 4. Here the net energy of system will be zero.6 = 24.5 J 2 Loss in PE between A and B = 1.5 + 19. S. S.8 ¥ 2 = 19.0 ¥ 62 = 18 J 2 Work done by friction = Change in energy = 18 – 24.1 J 1 But actual KE at B = ¥ 1. F – Mg = Ma 7 Mg fi . the floor also exerts force F on box in upward direction.3 j \ VD1M = 16 + 9 = 5 m/s en ga ge 2. 5q 2 q2 ( 2q ) 2 + = 2C1 2C2 16pe 0 R q2 q2 = 2C1 16pe 0 R .1 J 4.0 ¥ 3. 10Q + q -q q Fig. S. no impulse acts in the direction of motion. 38 NA = mg + P sin q NB = mg – P sin q NC = mg Clearly NA > NC > NB Now to find relation betwen frictions: (i) Let sliding does not occur: fA = P cos q. S.36 BD = BAC + BCB = m0 4p ˘ Èi ˘ m0 È i Í sin f1 ˙ + Í sin f2 ˙ p r 4 r Î1 ˚ Î 2 ˚ m0 È sin 60∞ sin 30∞ ˘ m 0i iÍ ˙= r2 ˚ 4p R 3 4p Î r1 7.37.20 dx dy = 11. r2 = 2R sin 60° = R 3 Force = mI = 2 (20 2 ) = 40 2 N 10. slope is negative. the water surface will make an q angle q with horizonB a tal. c. Ans (b): Current through the inductor before closing the switch = 10/(5 + 5) = 1 A Current through the inductor after closing the switch 20 (in steady state) I = = 4A 5 \ Df = L DI = 1.5 Wb 8.B. Cv = R. so velocity is positive. the molar specific heat at con3 5 stant volume. Ans (a. Alternately: T cos q + mg = B cos q fi (B – T) cos q = mg . S.. fC = P fi fC > fA = fB (ii) Let sliding occurs at all contact surfaces: fA = mNA. fC = mNC fi fA > fC = fB .30 Physics B A i 60° C r1 r2 30° R D Fig. Ans (a.(1) B sin q – T sin q = ma fi (B – T) sin q = ma . S. Ans (b): As shown in Fig. From A to C. fB = P cos q. because from O to B. acceleration is negative and from B to D. where tan q = a/g.. 2 2 For water vapour. acceleration is positive. 12. Cv = 3R Specific heat Cv of the mixture 3 5 2 ¥ R + 4 ¥ R + 1 ¥ 3R 16 2 2 = = R 2 + 4 +1 7 Pv t kq kq k (10Q + q ) + = 0 fi q = . Ans (d): For helium.. I y = = . for hydrogen. Ans (a): I x = = . fB = mNB. graph is opening downwards and from B to D graph is opening upwards. b. Fig. T The string will orient q mg itself perpendicular to the water surface. Ans (b): r1 = 2R sin 30° = R. Cv = R.L td . S. so velocity is negative. 6. inclination of –1 the string with vertical is tan (a/g) towards right.(2) From (1) and (2). c) C en ga ge Le ar ni ng Section B Resultant field: I = I x2 + I y2 = 20 2 16 23 R+R = R 7 7 a CP = Cv + R = di i .37 Hence. tan q = a/g dV dV 9.4Q R 2R 3R In V= fA P sin q P cos q NA mg P sin q P cos q fB NB (a) mg (b) P fA mg NC (c) Fig. From O to A and from C to D. slope is positive.20. d): From O to B.. S. S.1ˆ l1D 2m .39 1A + (b) ng 3. (d) is correct Section C (2) 5A+3A (a) (b) (c) ar ni After redrawing the circuit.00 W di In .1 Ê l2 D ˆ Ê 2m .1ˆ l1D = 2 ÁË d ˜¯ ÁË 2 ˜¯ d fi (2n – 1) 2 = 2m – 1 1 fi m = 2n 2 not possible because m and n are integers Let nth order minima of l2 coincides with mth order maxima of l1.Appendix B: Mock Tests B. R 2A + E1 E2 + 6.00 W 10 V R (3) (1) 4.31 14. Then Ê 2n . So particle (2) will be above the center of mass level.00 W 3. Ans (5): Since A.00 W 2A 3. then B Q x x P .00 W 3.00 W 10 V – 5A Fig.00 W E2 + 7 A Le 3. Then nl2 D ml1D = fi m = 2n d d Hence (b) is correct Let nth order minima of l2 coincides with mth order minima of l1. the following circuit can be redrawn as: R x B A x A P x x/2 x/2 A x/2 x x x x x x/2 Fig. Ans (b. Hence.1 fi 2n = =Á Ë 2 ˜¯ d d 2 fi 2m = 4n + 1 1 fi m = 2n + 2 Not possible because m and n are integers l D 2l D b 2 = 2 = 1 = 2b1 d a Let nth order maxima of l2 coincides with mth order maxima of l1.v2 cos q 2 ) =0 m+m So the horizontal velocity of center of mass is zero. center of mass will move in vertical direction. d): vcm. nl2 D Ê 2m .00 W 4. c. d) 13. Now v2 cos q2 = v1 cos q1 v cos q1 >1 fi 2 = (Q q2 > q1) v1 cos q 2 fi v2 > v1 fi v2 sin q2 > v1 sin q1 At any time.00 W ge 5A 4.E1 54 .00 A + 6. Then Pv t a mv1 cos q1 + m (. P.00 A 4.36 R = 2 = =9W 2 2 15. So.41 x x x B E .40 ga Current through 4 W = 5 A From loop (1): – 8(3) + E1 – 4(3) = 0 fi E1 = 36 V From loop (2): + 4(5) + 5(2) – E2 + 8(3) = 0 fi E2 = 54 V from loop (3): – 2R – E1 + E2 = 0 en C (c) 2n . b. Q and R are at same potential. d) (a) Let nth order maxima of l2 coincides with mth order minima of l1. Ans (a. Ans (a. we get E1 (d) Fig.L td .x = 16.1ˆ l2 D ml1D = Á Ë 2 ˜¯ D d fi m = 2 n – 1 Hence. particle (2) will be at greater height than particle (1). S. 3.42 Le v = – 80 cm I ¢ seen as object for the mirror 1 1 -1 fi + = fi v = .cos 2q ) c 20. 1.5 = 12.8 kV. Hence. Ans (d): V = pd2 l 4 . h Force due to one photon: f = 2 sin q l This force is perpendicular to the surface of cone.5 Å 4. 2 (la – lmin.5 Therefore.1) 18. the net force on the cone will be vertically downward.5 ¥ 0. Ans (5): Rate of mechanical work done = Fv where F is the force and v is the velocity. Ans (3): + = =0 v 60 • Pv t q a I¢ di 2 nE = IR fi n = pR2I/E By symmetry. net force on cone will be In I ar ni M ng A Fig. mgs (sin a – sin b) = = 1 1 Ê1 ˆ I w 2 + 2 Á mv 2 + I w 2 ˜ Ë2 ¯ 2 2 where w = v/r and I = mr2/2 Putting values and solving. Ans (d): Because of the thickness of the lens.4 kV and V 0.25 cm v -100 20 ga ge I ≤ will serve as a object for plane surface 1 (4 / 3) (1 .5 ¥ 4 = 50 W Work done in 10 s = 50 ¥ 10 = 500 J 4 / 3 1 ((3 / 4) .43 O I¢¢ 1. So. Ans (d): T = 2p . S. equivalent resistance between A and B L L < .24 ¥ 104 Å = 1 Å for 12. 1) = (la – lmin. Ans (2): Applying COE. Ans (c): lmin = 5. we cannot use the usual formulae of lens and a unique focal length cannot be derived.5 fi la = 1. power = 12. So time taken t = R Fig. v1 + v3 v3 Ê h ˆ F = nf sin q = n Á 2 sin q ˜ sin q Ë l ¯ p R2I (1 . Ê Blv ˆ F = Bil = B ◊ Á l Ë R ˜¯ = 19.5 Å for 24. S. so relative velocity between pulse and observer = v1 + v3.75 cm Paper 2 en Section A C M in any case K 2. Then q q = 0. Now. Ans (d): As the observe is moving. we get v = 2 m/s.32 Physics 21 21 x= r 10 40 17.L td .B. 2) fi 2la – 2 = la – 0. Ans (2): Power of light received by the cone = l (p R2).5 ¥ 0.(4 / 3)) = v 5 • v = +3. Let the number of photons hitting the cone per second is n. d) (s – b. F2 Æ second focus I Æ Image. DV 2Dd Dl 2 ¥ 0. Ans (a): v = w/K = 100 p /(0. c) (s – b. Hence. so it should be concave lens. d): When the current becomes constant. 16. (p – b. mR fi R = fi m=• (not possible) m -1 15. 19.5p) = 200 11.L td .5% V 6. Hence. 17. 7.1 1 = + = + = V d l 4 5 40 Pv t fi L ar ni Fig. d) (p) During AB: Temperature is constant (DU = 0) and pressure is increasing. c. tension.01 0. d) (q – a. then y < 2R. c. Ans (c): At x = 0 y = y1 + y2 = 2A cos 96pt cos 4 pt Sectiion C 18. Since the second focus is to the left of lens. d) (r – a.Appendix B: Mock Tests B. So also DQ <0 a= . Ans (a): w1 = 100 p w 92p f1 = = 50 Hz. c. cos 96pt = 0 or 4pt = 0 fi 96pt = (2n + 1)(p/2) and 4pt = (2m + 1)p/2 For 0 < t < 1. frequency etc. 12.5 2 2 Here n and m are integers. beam can be focussed before A. % error in volume = a L For y = 0. 1 1 .33 DV ¥ 100 = 2. So from PV = nRT.5 and . Hence. wavelength depends upon them. beam should be thin. Ans (c): Time period is not affected by a constant force acting along the line of SHM. Ans (b) 13. Ans (c): Use conservation of energy. respectively. than y = 2R mR fi 2 R = fi m=2 m -1 . l l Now L1 = L2 fi m 1 = n 2 fi m = 2n 2 2 For minimum. Ans (c) 14. If m > 2.< n < 95. v m 2 l1 l1 1 8.44 ng In F2 I The above is valid for paraxial rays. n = 1 fi m = 2 en Section B C 9. (p – b. d) Velocity at a point at distance x from the lower end: v= a= T = m m xg = xg m dv x -1/ 2 dx = g 2 dt dt g x -1/ 2 gx = g / 2 2 Velocity. PD across inductor will become zero. do not depend upon wavelength. But in turn. Ans (a. therefore. For thin beam to be focussed at A. Hence. S. Ans (a. Ans (a. di F1 Le F1 Æ first focus. the net amplitude becomes zero 100 times. Also energy stored in the inductor becomes constant if current through it becomes constant. Ans (a): mg is balanced by the electrostatic force. d) (q – a.= fi y= y • R m -1 If beam is focussed at A. d): 1 = = fi = v2 m1 l2 l2 2 ga ge l l L1 = m 1 and L2 = n 2 2 2 where m and n are the number of loops in L1 and L2. DW < 0.< m < 3. d): The incident rays parallel to the principal axis pass through (or appears to come from) second focus after refraction. c) (r – a. Ans (d) m 1 m -1 mR . the entire PD will be across R. y = R. volume decreases. f2 = = 46 Hz 2p 2p Beat frequency = f1 = f2 = 4 10. d) (s – c) (p) Flux entering through base: f = EA = EpR2 (q) Flux outgoing through curved part is same as the flux entering through base. so volume increases. flux through any closed surface is zero. . (p – a. So ÈÊ 3R ˆ 2 Ê R ˆ 2 ˘ Ep R 2 f = Ep ÍÁ ˜ . (r) During CD: Opposite to AB (s) During DA: Opposite to BC 20. (r) Flux through MNQP will be same as the flux through base from radius R/4 to 3R/4.B. d) (q – a.Á ˜ ˙ = Ë 4¯ ˙ 2 ÍÎË 4 ¯ ˚ R/4 3R/4 C en ga ge Le ar ni ng In di a Fig. Hence DW > 0.34 Physics (s) Net flux through the entire cone will be zero as in uniform electric field. S.L td . So also DQ > 0. d) (r – b.45 . P Q M N Pv t (q) During BC: Temperature increases (DU > 0) and pressure remains constant.